Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $r \neq 0$. $t = \dfrac{6}{7r(3r - 7)} \times \dfrac{21r^2 - 49r}{6} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $t = \dfrac{ 6 \times (21r^2 - 49r) } { 7r(3r - 7) \times 6 } $ $ t = \dfrac {6 \times 7r(3r - 7)} {6 \times 7r(3r - 7)} $ $ t = \dfrac{42r(3r - 7)}{42r(3r - 7)} $ We can cancel the $3r - 7$ so long as $3r - 7 \neq 0$ Therefore $r \neq \dfrac{7}{3}$ $t = \dfrac{42r \cancel{(3r - 7})}{42r \cancel{(3r - 7)}} = \dfrac{42r}{42r} = 1 $